The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+41 votes
7.9k views

Suppose the round trip propagation delay for a $10\text{ Mbps}$ Ethernet having
$48\text{-bit}$ jamming signal is $46.4\ \mu s$. The minimum frame size is:

  1. $94$
  2. $416$
  3. $464$
  4. $512$
asked in Computer Networks by Veteran (59.7k points)
edited by | 7.9k views
0
How Jamming Signal and Length of Packet is dependent?

Tt>=2*Tp

so, L>=2*Tp*Bandwidth

6 Answers

+59 votes
Best answer

The sender must be able to detect a collision before completely sending a frame.
So, the minimum frame length must be such that, before the frame completely leaves the sender any collision must be detected.

Now, the worst case for collision detection is when the start of the frame is about to reach the receiver and the receiver starts sending. Collision happens and a jam signal is produced and this signal must travel to the sender.

The time for this will be the time for the start of the frame to reach near the receiver $+$ time for the jam signal to reach the sender $+$ transmission time for the jam signal.

(We do not need to include transmission time for the frame as as soon as the first bit of the frame arrives, the receiver will have detected it). Time for the start of the frame to reach near the receiver $+$ Time for the jam signal to reach the sender $=$ Round trip propagation delay $= 46.4 \mu s$. So, 

$46.4 +\dfrac{48}{10}\text{(48 bits at 10 Mbps takes 4.8 micro sec.) = 51.2 μs.}$

Now, the frame length must be such that its transmission time must be
more than $51.2 \mu s$.

So, minimum frame length$= 51.2\times 10^{-6}\times 10\times 10^6=512\text{ bits}$.

http://gatecse.in/w/images/3/32/3-MACSublayer.ppt

A reference question from Peterson Davie:

43.  Suppose the round-trip propagation delay for Ethernet is 46.4 $\mu s$. This yields a minimum packet size of 512 bits (464 bits corresponding to propagation delay + 48 bits of jam signal).

(a)   What happens to the minimum packet size if the delay time is held constant , and the signaling rate rises to 100 Mbps?

(b)   What are the drawbacks to so large a minimum packet size?

(c)   If compatibility were not an issue, how might the specifications be written so as to permit a smaller  minimum packet size?

Another reference for requiring jam signal bits to be included for minimum frame size.

http://intronetworks.cs.luc.edu/current/html/ethernet.html

Can collision be detected by the source without getting the full jam signal (by change in current)?

Probably yes. But to be safe (from signal loss) the source waits for the entire jam signal. See below link

http://superuser.com/questions/264171/collisions-in-csma-cd-ethernet

answered by Veteran (369k points)
edited by
+4
Bhagwan ke liye Arjun sir ka peecha chhod do for this question :P
0
But Round Trip Time is the time the signal needs to come back to sender and when the signal came back to the sender it is obvious that the collision has occurred. Then why we need to consider jamming time also. I am not clear with this technique. Can anyone explain it?
+1
But sir jamming signal doesn't affect the minimum packet size.
+3

@Bad_Doctor

https://superuser.com/questions/264171/collisions-in-csma-cd-ethernet

Read this it should clear the concept.

0
But it is mentioned in Frozen that minimum frame transmission time is atleast two times the propagation delay.So is it partial statement?

Because looking at the discussion it seems min. frame transmission time is atleast two times the propagation delay+time to transmit jamming signal

@Arjun sir.Please confirm
+1

@rahul

frame transmission time is atleast two times the propagation delay+time to transmit jamming signal

Only in this question I have seen this, rest all questions

frame transmission time is atleast two times the propagation delay

But, in those questions there is no explicit mention of jamming signal.

+1

@arjun sir here i am unable to understand why the 96 is added for A to be start transmission ...

0
Must read from the link share by @VS Loyal to clear doubts
0

 

Try ro read the highlighted portion .

0
In Worst case

Now, the worst case for collision detection is when the start of the frame is about to reach the receiver and the receiver starts sending.The receiver would have transmitted some bits on link so. The NOISE(Twice the energy level as sent signal) generated when two signals collide will come to A in round trip time and it detects collision due to energy level sensing  so why it needs jam signal for detection of collision ?
+1 vote
ans c)
answered by Loyal (5.3k points)
+4

See here

We have to count the time for jamming signal also. So, it should be 512.

0
@Arjun sir, but jamming signals and data are propagated together in the cable no? why need to take them seperately? if jamming signal time is greater than collision detection time, then we have to consider them seperately no?
+12
First frame goes from A->B. Collision happens at B, jamming signal sent by B.

So, total time = 2 * propagation delay, one for frame and one for jam signal.

This is the time for the first bit of jamming signal to reach A. For A to identify collision all 48 bits must reach. And this time is the transmission time.
0
Arjun Sir, why sender looking all 48 bits of jam signal? To identify collision only need to jamming signal reach at the sender (that means it is enough to transmit a packet for minimum 2*propagation delay)
+2
How can it know that it is jamming signal unless whole 48 bits are received? To identify something is 'x' the whole 'x' must be received- unless there is some identification stuff- which isn't mentioned in the question.
0
Arjun Sir, There is a confusion in my mind that is when frame goes from sender A to receiver  B and collision happen at B then energy level in the channel at B is high and B is immediately sense that collision has occurred. After that B is sending a jam signal(i.e. 48 bit) and jam signal also have some different energy level . So this energy level whenever sense by A it knows that it is a jam signal immediately.

Sir,Why the A need to be sense all 48 bits . Why not it knows immediately just sensing the energy level like at receiver B?
0
How can the energy level be sensed by A?

Given 48 bits is the jamming signal. So, we can have a case where 47 bits are same as jamming signal and that can be part of a valid data.
+3
@Arjun, Issue here is that

Valid data can not come on Channel while we are transmitting data, while sending data, channel should sound Idle to sender i think.

Because no body else is sending data other than sender at that moment, even 1 bit coming from other direction means that collusion is happened !
0
are transmission time for frame and jamming signal same?as you have calculated transmission time for 48 bits where we needed to do for the frame?
0
if we consider a scenario in which A and B starts transmitting data packet at the same time. Then obviously collison will happen at the middle that is after half of Tp(propagation delay) time. Then collison signal will move to both A and B and when they detect collision again after half of Tp time, both will send jamming to inform other stations on the ethernet about the collision. So how can A and B depend on the jamming signal to detect the collision??
0 votes

The jam signal or jamming signal is a signal that carries a 32-bit binary pattern sent by a data station to inform the other stations of the collision and that they must not transmit.

Means sender already knows that collision has occurred than only he send Jamming signal to others.

Our motive here to let sender know about collision, which he already knew before sending Jamming signal.

So don't count Jamming signal time while calculating frame size.

Frame size must be = 208

answered by (27 points)
0 votes
Transmission delay>= 2*Prropagation delay

L/B.W=46.4 microsec

L=46.4 microsec*10Mbps

L=464b

note:-Use of Jamming signal:-If a collision is heard, both of the senders will send a jam signal over the Ethernet. This jam signal indicates to all other devices on the Ethernet segment that there has been a collision, and they should not send data onto the wire. After sending the jam signal, each of the senders will wait a random amount of time(decided by backoff algorithm) before beginning the entire process over. The random time helps to ensure that the two devices don't transmit simultaneously again.
answered by (33 points)
–2 votes
hey first read the question....i dunt know what you all are trying to do but simple solution is that-

to detect collision in a ethernet of 10Mbps bandwidth(which is general case) minimum frame size must be of 64 bytes or 512 bits and maximum is 1518bytes or 12144 bits (theoretically)

so answer (D) 512
answered by (475 points)
–5 votes

for a successful detection of collision it should be the case that :

$\begin{align*} \text{Transmission time} &\geq \text{RTT}\\ \frac{f+48}{10\times 10^6} &\geq 46.4\mu\\ f+48 &\geq 464\\ f &\geq 416 \end{align*}$

Minimum packet size = 416 bits.

answer = option B

answered by Boss (30.9k points)
Answer:

Related questions



Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

44,297 questions
49,785 answers
164,372 comments
65,857 users